I know that in general, a skew-symmetric matrix with indeterminate elements has a determinant that can be written as a square of some multivariable polynomial. How to prove this?

And if I do not know anything about the Pfaffian, can I prove the statement that a skew-symmetric matrix with integer entries has a determinant that is a square of some integer? I mean if someone knows how to prove it without using the Pfaffian?

1 Answer
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This is for the second part, a skew-symmetric matrix with integer entries

First, if $n$ is odd, then since $\det(A) = \det(A^T) = \det(-A) = (-1)^n \det(A)$, so $\det(A)= 0$, which is the square of an integer.

Now for $n$ even, we proceed by induction, and will show the statement is true over the rationals. Base case $n=2$ is obvious.

Inline Edit: Induction step. Suppose we want to show it for some $n=2k+2$. Consider the entries. We already know that the diagonals satisfy $a_{i, i} = 0$. If all the other entries are also 0, then we are done. Otherwise, WLOG we have $a_{1,2}\neq 0$. We will proceed to calculate the determinant of $A$. end edit

Using only row operations, where we add a rational linear multiple of the second row to the others, we can make the first column to be $(0, a_{2, 1}, 0, 0, \ldots, 0)^T$. Specifically, for row $k\neq 1, 2$, we have $b_{k, i} = a_{k, i} - \dfrac {a_{k, 1}}{a_{2, 1}} a_{2,i}$. Now, we use column operations, where we add a rational linear multiple of the second row to the others, and we can make the first row to be $(0, a_{1, 2}, 0, 0, \ldots, 0)$. Specifically, for column $k\neq 1, 2$, we have $c_{i, k} = b_{i, k} - \dfrac {b_{1, k}}{b_{1, 2}} b_{i, 2}$. Then, $\det(A) = a_{1, 2}^2 \det(C)$, and so it remains to show that $C$ is still a skew symmetric matrix.

@adamW I came up with this looking at the 4 by 4 case. I didn't feel that identifying 0's would help, since it doesn't really affect the general identity. In fact, there was no issue go dividing by 0, since we only divided by $a_{1, 2}$ and $a_{2, 1}$ which we can choose to be non-zero. Furthermore, with the definition of $B$, the 0 entries on the diagonal will have disappeared.
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Calvin LinJan 13 '13 at 18:37

You may have misunderstood, I only meant that I had to realize the zeros were there to understand your statement "expanding along that row will give us the result via the induction hypothesis"
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adam WJan 13 '13 at 18:53

@adamW Ah, I see. Hm, but I already stated that $a_{i,i}=0$. Will add a line for clarity.
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Calvin LinJan 13 '13 at 18:54

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+1 very nice. Just a comment: the entries of $C$ will not be integers but this is not a problem. If $r$ is a common denominator then $r^n$ is a square and $r^nC$ has integer entries.
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P..Jan 13 '13 at 20:25